zumdahl s chapter 11 n.
Skip this Video
Loading SlideShow in 5 Seconds..
Zumdahl’s Chapter 11 PowerPoint Presentation
Download Presentation
Zumdahl’s Chapter 11

Loading in 2 Seconds...

play fullscreen
1 / 27

Zumdahl’s Chapter 11 - PowerPoint PPT Presentation

  • Uploaded on

Zumdahl’s Chapter 11. Solutions. Solution Composition Concentrations H solution Hess’s Law undersea Solubilities Henry’s Law: Gases and Raoult’s Law Temperature Effects. Colligative Properties T BP Elevation T FP Depression Osmotic Pressure van’t Hoff Factor

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Presentation

Zumdahl’s Chapter 11

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
chapter contents
Solution Composition



Hess’s Law undersea


Henry’s Law: Gases

and Raoult’s Law

Temperature Effects

Colligative Properties

TBP Elevation

TFP Depression

Osmotic Pressure

van’t Hoff Factor

Colloids and Emulsions

Chapter Contents
solution composition
Solution Composition
  • Molarity, M = moles solute / liter sol’n.
    • Cannot be accurately predicted for mixtures because partial molar volumes vary.
    • If volumes don’t add, masses and moles do!
  • Molality, m = moles solute / kg solvent
    • Not useful in titration unless density known.
    • Useful in colligative effects.
  • Mole fraction, XA = moles A / total moles
conc of 50 by wt naoh
Conc. of 50% by wt. NaOH
  • Density at 20°C is 1.5253 g cm3
    •  each liter of solution weighs 1525.3 g
    • ½ that mass is NaOH, or 762.65 g
    • nNaOH = 762.65 g [ 1 mol/39.998 g ] = 19.067
    • [NaOH] = 19.067M but also
    • 19.067 mol / 0.76265 kg H2O = 25.001mand
    • nwater = 762.65 g [ 1 mol/18.016 g ] = 42.332
    • XNaOH =19.067 /(19.067+42.332)=0.31054
100 cc ea h 2 o c 2 h 5 oh
100 cc ea. H2O & C2H5OH
  • Want Proof? 50% by Volume 100 proof
  • Want Volume? Need densities!
    • At 20°C,  = 0.99823 & 0.79074 g/cc, resp.
    •  sol’n. mass = 99.823+79.074 = 178.897 g
    • By mass: 100%(79.074 / 178.897) = 44.201%
    • From tables:  = 0.92650 g/cc
    • V = 178.897 g/0.92650 g/cc = 193.09 cc
      •  It’s really 2 100cc / 193.09 cc = 103.58 proof
conceptual mixing enthalpies

Expand both solvent and solute

  • at the expense of H1 and H2
  • in lost intermolecular
  • interactions.
Conceptual Mixing Enthalpies
  • Merge the
  • expanded liquids
  • together recovering
  • H3 from the
  • new interactions.

But even if it

requires heat,

mixing may

well happen

since entropy

favors it!

3. If the exothermic

mixing exceeds

the endothermic

expansion, there

will be a net exo-

thermic heat of


underwater hess s law
Underwater Hess’s Law
      • Unrelated to basket weaving.
  • Since solutions are fluid, they need not expand then mix, requiring “upfront” $$.
  • Instead they acquire AB interaction as they lose AA and BB ones; pay as you go.
  • Hess doesn’t care; the overall enthalpy change$ will be the same.
  • It’s true that which of A or B is the solute or solvent is mere naming convention …
      • Which was the solute in that 50% cocktail?
  • Still solutes with low solubility are surely in the mole fraction minority.
  • And it is worthwhile asking what state parameters influence their solubility?
gas solubilities
Gas Solubilities
  • No doubt about it: pressure influences solubility. And directly.
      • CO2 in soft drinks splatter you with dissolution as you release the pressure above the liquid.
  • Henry’s Law codifies the relationship:
  • PA = kH•[A(aq)] (kH is Henry’s constant)
    • It applies only at low concentrations; so
    • It applies not at all to strongly soluble gases!
raoult s and henry s laws
Apply at opposite extremes.

Raoult when X~1

Henry when X~0

So Raoult to solvent and Henry to solute.

When XB is small, XB=[B]/55.51M for [water]=55.51M

Henry’s OK with X.









P = P° X

P kH’ X

Raoult’s and Henry’s Laws
raoult vs henry difference
Raoult vs. Henry Difference
  • When X~1, the solvent is not perturbed by miniscule quantities of solute. Solvent vaporization is proportional to solvent molecules at solution’s surface. Raoult
  • When X~0, solute is in an utterly foreign environment, surrounded only by solvent. kH reflects the absence of A-A interaction, and Henry applies.
solubility and temperature
Solubility and Temperature
  • Sometimes the AB interactions are so much weaker than AA or BB that A and B won’t mix even though entropy favors it.
    • Since T emphasizes entropy, some of the immiscible solutions mix at higher T.
  • Solidsolubilities normally rise with T.
    • Exceptions are known … like alkali sulfates.
gases flee hot solutions
Gases Flee Hot Solutions
  • You boiled lab water to drive out its dissolved gases, especially CO2.
    • That’s why boiled water tastes “flat.”
      • Genghis Khan invented tea (cha) to flavor the water his warriors refused to boil for their health as they conquered Asia and Eastern Europe.
  • Increased T expands Vgas, making it more favored by entropy vs. dissolved gas.
    • This time, no exceptions!
changed phase changes
The Phase Diagram

Mixing in a solute lowers solvent Pvapor

So TBP must rise.

Since the solvent’s solid suffers no Pvapor change, TFP must fall.

Liquid span must increase in solution.



Changed Phase Changes
elevating depressions
Elevating Depressions
  • Both colligative properties arise from the same source: Raoult’s Law.
  • Thermo. derivations of resulting T give:
    • Freezing Point Depression:
      • TFP = –Kf msolute where Kf~RTFP2 / Hfusion
    • Boiling Point Elevation:
      • TBP = +Kb msolute where Kb~RTBP2 / Hvap
  • Kf > Kb since Hfusion < Hvap
practical phase changes
Antifreeze / Summer Coolant are the same

Ethylene glycol (1,2-Ethanediol) is soluble in the radiator water, non-corrosive, nonscaling, and raises the boiling point in summer heat while lower-ing freezing point in winter.

“Road salt” is CaCl2 now since NaCl corrodes cars.

Practical Phase Changes
col liga tive utility
Colligative Utility
  • Ligare means “to bind.” These features are bound up with just numbers of moles.
      • NOT the identity of the molecules!
  • Indeed, Kf and Kb are seen not to depend on solute properties but on solvent ones.
    • So they’re used to count solute moles to convert weights to molar weights!
      • Not sensitive enough for proteins, MW~10 kg
exquisite sensitivity
Exquisite Sensitivity
  • To count protein moles, we need Osmotic Pressure that is very sensitive to [solute].
    • Solvent will diffuse across a membrane to dilute a concentrated solute solution.
    • If the solute is too large (protein!) to diffuse back, the volumemust increase.
    • Rising solution creates (osmotic) pressure to an equilibrium against further diffusion.
mw by osmotic pressure

MW by Osmotic Pressure, 
  • Thermodynamic derivation of the balance between  & diffusion on the equilibrium gives: V = nRT(!) or = MRT
    • E.g., 0.5 g in 50 cc yields 10 cm of pressure at 25°C (so RT = 24.5 atm L /mol)
    • 10 cm (1 ft/30.5 cm) (1 atm/33 ft) = .010 atm
    • [protein] =  / RT = 0.00041 mol/L
    • Wt = 0.5 g/0.05 L = 10 g/L MW = 24 kg
moles of what
Moles of What?
  • Doesn’t matter if property’s colligative.
  • Counts moles of ions if solute dissociates.
  • van’t Hoff Factor, i, measures ionization.
    • i multiplies molality in any of the colligative expressions to show apparent moles present.
      • It’s a stand-in for non-idealities too; pity.
    • So in 0.001mK3PO4, i should be nearly 4, and colligative properties see 0.004m?NO!
weak electrolyte corrections
Weak Electrolyte Corrections
  • PO43– is a conjugate base of HPO42–
    • Ka3 = 4.810–13so Kb1 = Kw/Ka3 = 0.021 for PO43– + H2O  HPO42– + OH–
    • Equilibrium lies to left, so start with [OH–] = [HPO42–] = 0.001–xand [PO43–] = x
    • (0.001–x)2 / x = 0.021 or x ~ 4.810–5 ~ 0
    • Counting K+, total moles ~ 0.003+2(0.001)
    • Soi ~ 0.005/0.001 = 5not 4. (4.95 with care)
reverse osmosis
Reverse Osmosis
    • If dilution across a semipermeable (keeps out solute) membrane builds pressure,
  • Pressure should be able to squeeze water back out of a solution! …if the membrane survives.
  • Desalination plants are critical in desert nations like the Gulf States & N. Africa.
      • Waste water is much more (salt) concentrated, an environmental hazard to local sea life unless ocean currents are swift enough to dilute it.
when is a solution not a solution
When is a SolutionNot a Solution?
      • When it’s a problem? 
  • Insoluble materials precipitate out of a solution at a rate that increases with their mass. So smallparticlesstay suspended.
  • With particle sizes of 1 m to 1 nm such suspensions are called colloids.
    • Since visible ~ 0.5m, the larger colloids scatter visible light efficiently! (Tyndall effect)
taxonomy of suspensions
Taxonomy of Suspensions

Dispersed Material Phase

Dispersing Medium Phase

aqueous colloids










Aqueous Colloids
  • Particles might be charged and stabilized (kept from coagulating) by electrostatics.
    • Even neutral ones will favor adjacency of one charge which develops double layer (an oppositely charged ionic shell) to stabilize the colloid.
    • “Salting out” destroys the colloid by over-whelming the repulsions with ionic strength.
      • Small, highly charged ions work best, of course.
surface chemistry liquids
Surface Chemistry (liquids)
  • Colloid study, a subset of surface science.
  • Colloid molecules must be insoluble in the dispersing medium.
    • Solubility governed by “like dissolves like.”
    • But surface tensions play a role as well since solutes display surface excess concentration.
      • Interfaces between phases are not simply at the bulk concentrations; influences segregation.
surface chemistry solids
Surface Chemistry (solids)
  • Industrial catalysts for many processes are solids.
    • Atoms and molecules adhere, dissociate, migrate, reassociate, and desorb.
    • Efficiency scales with catalyst surface area.
    • Area measured by adsorbing monolayers of gas (N2 ) and observing discontinuities as monolayer is covered.